Question

At a sudden contraction in a pipe the diameter changes from D 1 to D 2 . The pressure drop, Δ p , which develops across the contraction is a function of D 1 and D 2 , as well as the velocity, V , in the larger pipe, and the fluid density, rho, and viscosity, μ . Use D 1 , V , and μ as repeating variables to determine a suitable set of dimensionless parameters. Why would it be incorrect to include the velocity in the smaller pipe as an additional variable?

Answer 1

Answer:

Velocity in the smaller pipe should not be included as an additional variable.    

Explanation:

$\Delta P = f(D_1, D_2, V, \rho, \mu)$

The dimensional formula of the variables are

$\Delta P = FL^{-2} , D_1 = L, D_2=L, V=LT_{-1}, \rho =FL^{-4}T^2, \mu = FL^{-2}T$

Now using Buchingham's Pi Theorem, 6 - 3 = 3 dimensional parameters are required.

Use, D_1, V, \mu as the repeating variables.

Therefore, $\pi = \Delta PD_1^aV^b \mu^c$

$(FL^{-2})(L)^a(LT^{-1})^b(FL^{-2}T)^c = F^0L^0T^0$

From this

1+c=0

-2+a+b-2c=0

-b+c=0

c=-1, b = -1, a = 1

Now, $\pi_1=\frac{\Delta PD_1}{V\mu}=\frac{(ML^{-1}T^{-2})L}{(LT^{-1})(ML^{-1}T^{-1})}=M^0L^0T^0$

For $\pi_2 = D_2D_1^aV^b\mu^c$

$F^0L^0T^0=L(L)^a(LT^{-1})^b(FL^{-2}T)^c$

c = 0

1 + a + b - 2c = 0

-b + c = 0

Therefore, a = -1, b = 0, c = 0

$\pi_2 = \frac{D_2}{D_1}$

For $\pi_3$

$\pi_3 = \rho, D_1^a V^b\mu^c$

$F^0L^0T^0 = (FL^{-4}T^2)(L)^a(LT^{-1})^b(FL^{-2}T)^c$

1 + c = 0

-4 + a + b - 2c = 0

2-b+c=0

c=-1, b=1, a = 1

Therefore, $\pi_3 = \frac{\rho D_1V}{\mu}$

Now checking,

$\pi_3 = \frac{(ML^{-3})(L)(LT^{-1})}{ML^{-1}T^{-1}} = M^0L^0T^0$

Therefore, $\frac{\Delta P D_1}{V \mu} = \phi (\frac{D_2}{D_1}, \frac{\rho D_1 V}{\mu})$

From continuity equation

$V\frac{\pi}{4}D_1^2 = V_s \frac{\pi}{4}D_2^2$

$V_s = V (\frac{D_1}{D_2})^2$

$V_s$ is not independent of $D_1,D_2, V$

Therefore it should not be included as an additional variable.